## Trending News

# calculus: find the limit using limit laws and/or algebra?

find the limit as x approaches 3

(2x^2-6x)/(2x^2-5x-3)

please show the algebra

### 6 Answers

- s kLv 79 years agoFavorite Answer
lim_{x->3} (2x^2 - 6x)/(2x^2 - 5x - 3)

= lim_{x->3} 2x(x - 3)/(2x^2 - 6x + x - 3)

= lim_{x->3} 2x(x - 3)/(2x(x - 3) + (x - 3))

= lim_{x->3} 2x(x - 3)/((x - 3)(2x + 1))

= lim_{x->3} 2x/(2x + 1)

= 6/7

- 9 years ago
The problem can be solved as follows:

L(x->3) [(2x^2 -6x)/(2x^2 -5x -3)]

= L(x->3) [2x(x -3)/(2x^2 -6x + x -3)]

= L(x->3) [2x(x -3)/(2x(x-3) + 1(x -3))]

= L(x->3) [2x(x -3)/((x-3)(2x + 1))]

= L(x->3) [2x/((2x + 1))] (Cancelling out x-3 from both numerator and denominator is allowed as x tends to 3 but is not equal to 3)

= 6/7

The solution is more easily obtainable using the L'Hôpital's rule. Applying L'Hôpital's rule for the problem at hand is possible since the conditions for its application are satisfied:

1) With numerator f(x) = 2x^2 -6x and g(x) = 2x^2 -5x -3, f and g is differentiable on an interval L = (2.9,3.1)

2) L(x->3) f(x) = 0 and L(x->3) g(x) = 0

3) L(x->3) f'(x)/g'(x) exists

4) g'(x) ~= 0 for all x in L

(See http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_... for reference).

The answer then is:

L(x->3) [(2x^2 -6x)/(2x^2 -5x -3)]

= L(x->3) [(4x -6)/(4x -5)]

=6/7

- 9 years ago
The numerator factors to 2x(x-3) and the denominator factors to (x-3)(2x+1).

So you have 2x(x-3)/(x-3)(2x+1).

The (x-3)'s reduce to one and you're left with

The limit as x approaches 3 of 2x/(2x+1).

Plug in 3 for x and you get 6/7. Viola!

- Kali PrasadLv 69 years ago
first factor

2x^2-6x = 2x(x-3)

2x^2 - 5x - 3 = (2x+1)(x-3)

divide numerator and denominator by common factor to get 2x/(2x+1) which is 6/7 when x =34

- How do you think about the answers? You can sign in to vote the answer.
- Anonymous9 years ago
factor 2x out of top:

2x(x-3) / (2x^2-5x-3)

factor the denominator:

2x(x-3) / (2x+1)(x-3)

(x-3) cancels out of top and bottom:

2x / 2x+1

plugging in 3 for x gives 6/7 = limit as x approaches 3

Hint: for help next time you can go to wolframalpha.com

- Anonymous9 years ago
lim(x→3) (2x²-6x)/(2x²-5x-3)=lim(x→3) 2x(x-3)/(2x+1)(x-3) =lim(x→3) 2x/(2x+1)

SO

lim(x→3) (2x²-6x)/(2x²-5x-3) → 2(3)/2(3)+1

→ 6/7